Question

An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive xx direction and has a magnitude of 6.9 N; a second force has a magnitude of 4.5 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

Answer 1

Answer with Explanation:

We are given that

`F_1=6.9 N`

`F_2=4.5 N`

We have to find the direction and magnitude of the third force acting on the object.

Resultant force,F=
`√(F^2_1+F^2_2)`

`F=√((6.9)^2+(4.5)^2)=8.24 N`

The object moves with constant velocity .Therefore, net force on object is equal to zero

So,Third force,
`F_3=F=8.24 N`

Direction,
`\theta=tan^(-1)((F_2)/(F_1))`

`\theta=tan^(-1)((4.5)/(6.9))=33.02^(\circ)`

Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.

Therefore,
`\theta=\pi-\theta=180-33.02=146.98^(\circ)`